Question

Determine the convergence or divergence of the sequence an=(3n^2-n+4)/(2n^2+1) and if the sequence converges, find its limit?

Answer 1

Converges to `3/2.`

Explanation:

A sequence `a_n` converges if `lim_(n->oo)a_n=L ne +-oo,` IE, for large values of `n, a_n` approaches some finite value and does not grow or decrease without bound.

Consider `a_n=(3n^2-n+4)/(2n^2+1)`

`lim_(n->oo)(3n^2-n+4)/(2n^2+1)=lim_(n->oo)((3n^2-n+4)/n^2)/((2n^2+1)/n^2)`

We divide the entire limit by the highest exponent in the denominator, `n^2`.

Simplify:

`lim_(n->oo)(3-1/n+4/n^2)/(2+1/n^2)=(3-1/oo+4/oo)/(2+1/oo)=3/2`

So, the sequence converges to `3/2.`